As shown in the figure, a particle of mass m and another particle of mass 10m are connected with a string. Friction is sufficient to prevent the slipping of 10 m. Mass mis given a velocity u in vertical direction. For complete circular motion of mass m :-

Let say speed at top position is v.Using energy conservation for smaller mass, Ui+Ki=Uf+Kf​⇒0+12mu2=mgL+12mv2​⇒v2=u2−2gL............1At top position using FBD we can write,T+mg=mv2L​⇒T=mv2L−mgCondition 1 : Tension should not become zero at top positionT>0​⇒v2>gL​⇒u2−2gL>gL​⇒u>3gLCondition 2 : The larger mass should not break contact with groundT<10mg​⇒mv2L−mg<10mg​⇒v2<11gL​⇒u2−2gL<11gL​⇒u<13gLThus, 3gL